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What does bar represent on a differentiation? I had an equation like this

$$\frac{d}{dt}(\sum_i p_i\cdot r_i)=2T+\sum_i F_i \cdot r_i$$

Now they said to integrate it.

The time average of Eq. (3.24) over a time interval τ is obtained by integrating both sides with respect to t from 0 to τ , and dividing by τ :

Then I can see a bar on top of above equation.

$$\frac{1}{\tau}\int_0^\tau \frac{dG}{dt}\equiv\frac{\overline{dG}}{dt}=\overline{2T}+\overline{\sum_i F_i \cdot r_i}$$

Does the bar represent integration? I didn't ask why there's $\frac{1}{\tau}$ cause the book says they had divided by $\tau$ but my question is if we had divided by $\tau$ then why there's no $\tau$ in second and third term? (still I have the question)

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    $\begingroup$ Bar means averaging here (averaging by time). $\endgroup$ Sep 22, 2021 at 9:40
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    $\begingroup$ Time average of any quantity $A$ is $\bar{A}=\frac{1}{\tau}\int_0^\tau A dt$. Averages properties follow from integration properties. $\endgroup$ Sep 22, 2021 at 9:42

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Just as the AVERAGE of n numbers is the sum of the numbers divided by n so the AVERAGE of a function defined on the interval from a to b is the integral of the function from a to b divie by b- a.

That is what this is saying. As others have said, the over-line indicates the AVERAGE.

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  • $\begingroup$ I didn't ask why there's 1/τ cause the book says they had divided by τ but my question is if we had divided by τ then why there's no τ in second and third term? (still I have the question) $\endgroup$
    – Unknown
    Sep 22, 2021 at 13:43
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    $\begingroup$ The integration is from 0 to $\tau$ so they are dividing by $\frac{1}{\tau}- 0$ to calculate the averag (the "b- a" in what I wrote before. $\endgroup$
    – user247327
    Sep 22, 2021 at 15:43

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